 Chapter 2 Waveguide.

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Chapter 2 Waveguide

Chapter Outlines Chapter 2 Waveguide
Rectangular Waveguide Fundamentals Waveguide Field Equations Parallel Plate Waveguide Rectangular Waveguide Modes Cylindrical Waveguide Fundamentals Cylindrical Waveguide Modes Resonant Cavity Dielectric Waveguide

Introduction WAVEGUIDE  any structure that supports propagation of a wave. In general usage: The term waveguide refers to constructs that only support non TEM mode propagation, name in the TE and TM Mode. It also unable to support wave propagation below a certain frequency, or cutoff frequency

Introduction (Cont’d..)
A waveguide is another means of guiding the EM energy from one point to another (same as transmission line). Some differences between waveguide and transmission line (TLine) : TLine can only support TEM wave whereas waveguide can support many possible field configurations.

Introduction (Cont’d..)
At microwave frequencies (3 to 300 GHz), TLine becomes inefficient due to skin effect and dielectric losses, but waveguides are used at microwave frequencies to obtain larger bandwidth and lower signal attenuation. TLine can operate above dc (f =0) to a very high frequency, but waveguide can operate only above cutoff frequency and therefore acts as a high pass filter.

Introduction (Cont’d..)
The most common waveguide types:

1.1 Rectangular Waveguide Fundamentals
A cross section of rectangular waveguide is shown below: Propagation is in the +z direction or out of page. Conducting walls  brass, copper or aluminium. Chosen to be thick enough for mechanical rigidity and several skin depths over the frequency of interest. The inside wall  electroplated with silver or gold to improve performance

Rectangular Waveguide Fundamentals (Cont’d..)
The interior dimensions are a x b, where the longer side is a. ‘a’ dimension: Determines the frequency range of the dominant, or lowest order, the mode of propagation. Usually operates in lowest propagating mode, since higher order  higher attenuation + difficult to extract from guide. ‘b’ dimension: Affects attenuation, smaller b has higher attenuation. Also sets the max power capacity Usually half of the ‘a’ dimension

Rectangular Waveguide Fundamentals (Cont’d..)
Waveguide can support TE and TM modes, where: In TE Modes, the electric field is transverse to the direction of propagation. Some magnetic field component in the direction of propagation. In TM Modes, the magnetic field is transverse and an electric field component must be in the propagation direction.

Rectangular Waveguide Fundamentals (Cont’d..)
The order of the mode refers to the field configuration in the guide and is given by ‘m’ and ‘n’ integer subscripts, as TEmn and TMmn. The ‘m’ subscript corresponds to the number of half wave variations of the field in x direction The ‘n’ subscript corresponds to the number of half wave variations of the field in y direction

Rectangular Waveguide Fundamentals (Cont’d..)
In conjunction with the guide dimensions, m and n determine the cutoff frequency for a particular mode. For conventional rectangular waveguide filled with air, where a = 2b, the dominant or lowest order mode is TE10 with cutoff frequency fc10 = c/2a

Rectangular Waveguide Fundamentals (Cont’d..)
The relative cutoff frequencies for the first 12 modes of conventional rectangular waveguide filled with air, Location of modes relative to the dominant TE10 mode in standard rectangular waveguide where a=2b.

Rectangular Waveguide Fundamentals (Cont’d..)

Example 1 Calculate the cutoff frequency for the first four modes of WR284 waveguide.

Solution to Example 1 From table, the guide dimensions are a=2.840 inches and b=1.340 inches. Converting to metric units: Therefore, where

Solution to Example 1(Cont’d..)
Then, we have: This agrees with the cutoff frequency cited in table. Then : not same with fc10 since a ≠ 2b

Solution to Example 1(Cont’d..)
and for the fourth mode,

Rectangular Waveguide Fundamentals (Cont’d..)
The field pattern for two modes where E only varies in the x direction, since n=0, the field is constant in the y direction. The field patterns and associated field intensities in a cross section of rectangular waveguide for (a) TE10 and (b) TE20. Solid lines indicate electric field; dashed lines are the magnetic field.

1.2 Waveguide Field Equations
Beginning with Maxwell’s equations, develop the time harmonic field equations for rectangular waveguide. For simplicity, consider the guide filled with lossless, charge free media and the walls to be perfect conductors.

Waveguide Field Equations (Cont’d..)
For conventional rectangular waveguide, the field components in Cartesian coordinates are: Inserting these equations into previous Maxwell’s equation..

Waveguide Field Equations (Cont’d..)
From previous divergence equations,

Waveguide Field Equations (Cont’d..)
From previous wave equation,

Waveguide Field Equations (Cont’d..)
From the first expanded Maxwell’s equations,

Waveguide Field Equations (Cont’d..)
Then, consider the fields only propagate in the z direction, in harmonic fields: The partial derivative with respect to z is: Substitute these into the expanded Maxwell’s equations

Waveguide Field Equations (Cont’d..)
Those equations can be reduced to:

Waveguide Field Equations (Cont’d..)
Wave equations also can be reduced to: With,

Waveguide Field Equations (Cont’d..)
So, using these equations, we can find expression for the four transverse components (Ex, Ey, Hx, Hy) in terms of z directed components (Ez and Hz), where: (1) (2)

Waveguide Field Equations (Cont’d..)
And also.. (3) (4) Try to derive these four equations on your own!

Waveguide Field Equations (Cont’d..)
So, these four important equations will be used to find the transverse components for TM and TE mode, where: for TM mode, Hz=0, then use these four equations to find the transverse components. for TE mode, Ez=0, then use these four equations to find the transverse components.

Waveguide Field Equations (Cont’d..)
for TM mode, Hz=0, equation (1) to (4) can be reduced to: (6) (5) (7) (8)

Waveguide Field Equations (Cont’d..)
for TE mode, Ez=0, equation (1) to (4) can be reduced to: (10) (9) (11) (12)

1.3 Parallel Plate Waveguide
The parallel plate waveguide is the simplest type of guide that can support TM and TE Modes, and TEM as well because it’s formed from two plates as shown. The width W is assumed to be much greater than the separation d, so that fringing fields and any x variation can be ignored. A material with permittivity εr and permeability µr is assumed to fill the region between the two plates.

Parallel Plate Waveguide (Cont’d..)
By considering the boundary condition, the magnitude E will change with y, E=0 at y=0 and at y=a but maximum in the middle. Magnitude E will not change with x since x is infinity (no boundary), so the value will constant . The E will along the propagation i.e. +z direction. Thus, y=a z y=0 x

Parallel Plate Waveguide Modes (Cont’d..)
For TM Mode For TM Mode, the magnetic field has its components transverse or normal to the direction of wave propagation, where Hz=0 and a nonzero Ez fields. So, Since the field is gong to +z direction, it can be reduced to only:

Parallel Plate Waveguide TM Modes (Cont’d..)
With , solution for this is: At y=0 Ezs =0 , thus Hence B=0 At y=a Ez =0 This valid if or n=1,2,3,4….

Parallel Plate Waveguide TM Modes (Cont’d..)
Thus with , Then, from equation (5) to (8), since d/dx=0, so only (5) and (7) remain, therefore:

Parallel Plate Waveguide Modes (Cont’d..)
For TE Mode For TE Mode, the electric field has its components transverse or normal to the direction of wave propagation, where Ez=0 and a nonzero Hz fields. So, Since the field is gong to +z direction, it can be reduced to only:

Parallel Plate Waveguide TE Modes (Cont’d..)
With , solution for this is: But, the boundary are that Ex=0 at y=0,a. So that from (10), At y=0 Ex =0 , thus Hence A=0

Parallel Plate Waveguide TE Modes (Cont’d..)
At y=a Ex =0 This valid if or n=1,2,3,4…. Thus with , and get back the solution for Hz,

Parallel Plate Waveguide TE Modes (Cont’d..)
Then, from equation (9) to (12), since d/dx=0, so only (10) and (12) remain, therefore:

1.4 Rectangular Waveguide Modes
For TM Mode For TM Mode, the magnetic field has its components transverse or normal to the direction of wave propagation. At the walls of the waveguide, the tangential components of the E field must be continuous, that is:

Rectangular Waveguide TM Modes (Cont’d..)
For TM mode, Hz=0, then lets: Where X is a function of x and Y is a function of y Then the wave equation becomes:

Transitional Page Rectangular Waveguide TM Modes (Cont’d..)
We divide the equation by XY Transitional Page and then Let’s or and

Rectangular Waveguide TM Modes (Cont’d..)
Solution for both equations: And the whole expression becomes:

Transitional Page Rectangular Waveguide TM Modes (Cont’d..)
We know that the tangential electric fields at the walls of the waveguide must be zero. Then by applying the boundary conditions: Applying boundary condition at x=0 where Ez=0 Transitional Page So C1=0 Applying boundary condition at y=0 where Ez=0 So D1=0

Rectangular Waveguide TM Modes (Cont’d..)
Then the equation reduced to Applying the other boundary condition x=a where Ez =0 E0 and sin kyy cannot be zero This means that sin kxa = mπ = 0 or where m=0,1,2,3,4…

Transitional Page Rectangular Waveguide TM Modes (Cont’d..)
Applying the other boundary condition y=b where Ez =0 Transitional Page But E0 and sin kxx cannot be zero, This means that sin kyb = 0 or where n =0,1,2,3,4… So,

Rectangular Waveguide TM Modes (Cont’d..)
Then, substitute Hzs=0 and we have:

Transitional Page Rectangular Waveguide TM Modes (Cont’d..)
and also.. Transitional Page Denote as TMmn ,the field vanish for TM00 , TM10 and TM01. The lowest mode is TM11

Rectangular Waveguide TM Modes (Cont’d..)
Useful relations to be remember, Wave number, The propagation constant is: where

Transitional Page Rectangular Waveguide TM Modes (Cont’d..)
Where we have 3 possibilities depending on k (or ω), m and n: Cutoff mode : Transitional Page At this time, ω is called cutoff angular frequency:

Rectangular Waveguide TM Modes (Cont’d..)
Evanescent mode : We have no propagation at all. These non propagating or attenuating modes are said to be evanescent. Propagating mode :

Transitional Page Rectangular Waveguide TM Modes (Cont’d..)
Where the phase constant becomes: For each mode (combination of m and n) thus has a cutoff frequency fcmn given by: Transitional Page The cutoff frequency is the operating frequency below which attenuation occurs and above which propagation takes place.

Rectangular Waveguide TM Modes (Cont’d..)
The cutoff wavelength, The phase constant, The intrinsic wave impedance, Where, Intrinsic impedance in the medium

Rectangular Waveguide Modes (Cont’d..)
For TE Mode For TE Mode, the electric field has its components transverse or normal to the direction of wave propagation. At the walls of the waveguide, the tangential components of the E field must be continuous, that is:

Transitional Page Rectangular Waveguide TE Modes (Cont’d..)
For TE mode, Ez=0, then lets: Where X is a function of x and Y is a function of y Transitional Page From previous derivation, the wave equation becomes:

Rectangular Waveguide TE Modes (Cont’d..)
Hz cannot impose boundary condition since it is not zero at boundary, so we determine Ex and Ey as follow : From, it becomes: Then reduced to:

Transitional Page Rectangular Waveguide TE Modes (Cont’d..) From,
it becomes: Then reduced to: Transitional Page

Rectangular Waveguide TE Modes (Cont’d..)
Then by applying the boundary conditions: Applying boundary condition at y=0 where Ex=0 So D2=0, then applying boundary condition at x=0 where Ey=0 So C2=0

Transitional Page Rectangular Waveguide TE Modes (Cont’d..)
Applying the other boundary condition x=a where Ey =0 This means that sin kxa = 0 or where m=0,1,2,3,4… Transitional Page Applying the other boundary condition y=b where Ex =0 This means that sin kyb = 0 or where n=0,1,2,3,4…

Rectangular Waveguide TE Modes (Cont’d..)
So, So, Then, substitute Ez=0 and we have:

Transitional Page Rectangular Waveguide TE Modes (Cont’d..)
and also.. Transitional Page Denote as TEmn ,the field vanish for TE00. The lowest mode is TE10 for a>b and TE01 for b>a.

Rectangular Waveguide Modes (Cont’d..)
From equations for the TM and TE modes, we can obtain the field patterns.. For example, the dominant TE10 mode, where m = 1 and n = 0, so the Ex, Ey, Hx, Hy and Hz equations becomes, So then in the time domain,

Rectangular Waveguide Modes (Cont’d..)
Similarly..

Rectangular Waveguide Modes (Cont’d..)
So, for TE10 mode, the variation of the E and H fields with x in an xy plane, say plane cos (ωt-βz)=1 for Hz and plane sin (ωt-βz)=1 for Ey and Hx

Rectangular Waveguide Modes (Cont’d..)
The corresponding field lines:

Rectangular Waveguide Modes (Cont’d..)
Field lines for some of the lower order modes of a rectangular waveguide :

Example 2 A rectangular waveguide with dimension a=2.5 cm, and b=1 cm is to operate below 15.1 GHz. How many TE and TM modes can the waveguide transmit if the guide is filled with medium characterized by σ=0, µr=1, ε=4ε0. Calculate the cutoff frequencies of the modes.

Solution to Example 2 The cutoff frequency is given by:
With a = 2.5b or a/b = 2.5, So, Or,

Solution to Example 2 (Cont’d..)
Since we are looking for cutoff freq below 15.1 GHz, a systematic way is to fix m or n and increase the other until fcmn is greater than 15.1 GHz. So, by fixing m and increasing n, Thus, for fcmn < 15.1 GHz, the maximum n = 2.

Solution to Example 2 (Cont’d..)
Then, fix n and increase m, Thus, for fcmn < 15.1 GHz, the maximum m = 5.

Solution to Example 2 (Cont’d..)
We know the maximum value of m and n, so try other possible combinations in between the maximum values.

Solution to Example 2 (Cont’d..)
Those modes whose cutoff freq are less or equal to 15.1 GHz will be transmitted, that is 11 TE modes and 4 TM modes, as illustrated below:

Example 3 In a rectangular waveguide with dimension a=1.5 cm, b=0.8 cm, σ=0, µ=µ0, ε=4ε0, find: The mode of operation The cutoff frequency The phase constant The propagation constant The intrinsic wave impedance

Solution to Example 3 We could find that the given expression is in instantaneous field expression form which obtained from the phasor forms by using: From the given expression we could find that m=1, and n=3. That is the guide is operating at TM13 or TE13. Suppose that after this, we choose TM13 mode.

Solution to Example 3 (Cont’d..)
Where for air filled waveguide, Hence, for m=1 and n=3, the cutoff frequency is:

Solution to Example 3 (Cont’d..)
The phase constant, Where, or

Solution to Example 3 (Cont’d..)
The propagation constant, Because it’s in propagating mode, So, The intrinsic wave impedance,

Transitional Page 1.5 Cylindrical Waveguide Fundamentals
A hollow metal tube of circular cross section also supports TE and TM waveguide modes as shown. Transitional Page

Cylindrical Waveguide Fundamentals (Cont’d..)
For cylindrical waveguide, the field components is in cylindrical coordinates which are: Inserting these equations into previous Maxwell’s equation,

Transitional Page Cylindrical Waveguide Fundamentals (Cont’d..) (13)
By using the same method of derivation for rectangular waveguide (starting from slide 21), we could get four equations of Eρ, Eφ, Hρ and Hφ in terms of Ez and Hz Transitional Page (13) (14) (15) (16) Try this!!!!

1.6 Cylindrical Waveguide Modes
For TM Mode For the TM modes of the circular waveguide, we must solve Ez from the wave equation in cylindrical coordinates, then through a very long and difficult derivation, we could get the transverse fields as:

Cylindrical Waveguide TM Modes (Cont’d..)
and also.. Transitional Page

Cylindrical Waveguide TM Modes (Cont’d..)
Some useful parameters for TM mode: Wave impedance for TM modes Cutoff wavenumber Propagation constant of the TMnm modes Cutoff frequency

Transitional Page Cylindrical Waveguide TM Modes (Cont’d..)
Values of pnm for TM Modes of a Circular Waveguide n pn1 pn2 pn3 2.405 5.520 8.654 1 3.832 7.016 10.174 2 5.135 8.417 11.620 Transitional Page pnm is the roots of Jn(x) which recognized as solution for Bessel’s differential equation.

Cylindrical Waveguide Modes (Cont’d..)
For TE Mode For the TE modes of the circular waveguide, we must solve Hz from the wave equation in cylindrical coordinates, after which we could get the transverse fields as:

Cylindrical Waveguide TE Modes (Cont’d..)
and also.. Transitional Page

Cylindrical Waveguide TE Modes (Cont’d..)
Some useful parameters for TE mode: Wave impedance for TE modes Cutoff wavenumber Propagation constant of the TEnm modes Cutoff frequency

Transitional Page Cylindrical Waveguide TE Modes (Cont’d..)
Values of p’nm for TE Modes of a Circular Waveguide n p'n1 p’n2 p’n3 3.832 7.016 10.174 1 1.841 5.331 8.536 2 3.054 6.706 9.970 Transitional Page p’nm is the roots of Jn(x) which recognized as solution for Bessel’s differential equation.

Field lines for some of the lower order modes of a cylindrical waveguide :

1.7 Resonant Cavity The length of resonator, d is made multiple of waveguide wavelength, i.e. p =1,2,3…… Resonator wavelength can be calculated as : where

Resonant Cavity (Cont’d..)
And kc for rectangular waveguide For cylindrical wave guide TM mode TE mode

Example 4 A cylindrical resonator has a radius of 5cm which is used to measure frequency from 8GHz to 12GHz at TE11 mode . What is the required length, d for tuning those frequency in that particular mode.

Solution to Example 4 First we calculate the cutoff wavelength
First frequency wavelength at 8 GHz, Second frequency wavelength at 12 GHz,

Solution to Example 4 (Cont’d..)
Calculate the length of wave guide For first frequency at 8 GHz, For second frequency at 12 GHz,

Solution to Example 4 (Cont’d..)
So, the cavity need to have length, d in this range in order to make the cavity operates at resonant frequencies between 8GHz to 12 GHz:

Dielectric filled wave guide
1.8 Dielectric Waveguide Air filled wave guide Dielectric filled wave guide

Chapter 2 Waveguide End